# Proof of Rank–nullity theorem

I'm trying to understand the proof of Rank–nullity theorem,but there are parts that I don't understand:

Steinitz exchange lemma

If $${\displaystyle U=\{u_{1},\dots ,u_{m}\}}$$ is a set of $${\displaystyle m}$$ linearly independent vectors in a vector space $${\displaystyle V}$$, and $${\displaystyle W=\{w_{1},\dots ,w_{n}\}}$$ span $${\displaystyle V}$$, then:

1. $${\displaystyle m\leq n}$$;
1. There is a set $${\displaystyle W'\subseteq W}$$ with $${\displaystyle |W'|=n-m}$$ such that $${\displaystyle U\cup W'}$$ spans $${\displaystyle V}$$.

Rank–nullity theorem

Let $${\displaystyle V}, {\displaystyle W}$$ be vector spaces, where $${\displaystyle V}$$ is finite dimensional. Let $${\displaystyle T\colon V\to W}$$ be a linear transformation. Then

$${\displaystyle \operatorname {Rank} (T)+\operatorname {Nullity} (T)=\dim V}$$

Proof

Let $${\displaystyle V,W}$$ be vector spaces over some field $${\displaystyle \mathbb {F} }$$ and $${\displaystyle T}$$ defined as in the statement of the theorem with $${\displaystyle \dim V=n}$$.

As $${\displaystyle \operatorname {Ker} T\subset V}$$ is a subspace, there exists a basis for it. Suppose $${\displaystyle \dim \operatorname {Ker} T=k}$$

I know that $$\text{ker T}$$ is a subset of a finite set $$V$$,and hence is finite,but how does this imply that $$\text{ker T}$$ does have a basis?

and let $${\displaystyle {\mathcal {K}}:=\{v_{1},\ldots ,v_{k}\}\subset \operatorname {Ker} (T)}$$ be such a basis.

We may now, by the Steinitz exchange lemma, extend $${\displaystyle {\mathcal {K}}}$$ with $${\displaystyle n-k}$$ linearly independent vectors $${\displaystyle w_{1},\ldots ,w_{n-k}}$$ to form a full basis of $${\displaystyle V}$$.

Let

$${\displaystyle {\mathcal {S}}:=\{w_{1},\ldots ,w_{n-k}\}\subset V\setminus \operatorname {Ker} (T)}$$ such that

$${\displaystyle {\mathcal {B}}:={\mathcal {K}}\cup {\mathcal {S}}=\{v_{1},\ldots ,v_{k},w_{1},\ldots ,w_{n-k}\}\subset V}$$ is a basis for $${\displaystyle V}$$.

What is $$V,W,W',U$$ here? Since the proof uses Steinitz exchange lemma,however I can't recognize $$V,W,W',U$$.

From this, we know that $${\displaystyle \operatorname {Im} T=\operatorname {Span} T({\mathcal {B}})=\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} T({\mathcal {S}})}$$

Why $$\text{Im}\; T=\text{span} \;T (\mathcal B)$$?

And why $$\operatorname {Span} \{T(v_{1}),\ldots ,T(v_{k}),T(w_{1}),\ldots ,T(w_{n-k})\}=\operatorname {Span} \{T(w_{1}),\ldots ,T(w_{n-k})\}$$?

• Every finite-dimensional space has a basis. math.stackexchange.com/questions/2785769/… Commented Dec 27, 2020 at 17:27
• Second question: $V,W$ were given as your vector spaces, $W'$ is the $n-k$ linearly independent vectors $w_1,\dots,w_{n-k}$, $U$ is $\mathcal K$. Third question: Im $T$ is Span$T(\mathcal B)$ by definition; the image of a linear map is all vectors spanned by a basis of its image. Fourth question: it is because $v_1,\dots,v_k$ are in the kernel, so they don't "change" the span. Commented Dec 27, 2020 at 17:32

1. $$\ker T$$ is a subspace, hence it is a vector space, so it has a basis.

2. $$V$$ is $$V$$, $$W$$ is any fixed basis of $$V$$, $$U=\mathcal K$$ and $$W'=\mathcal S$$.

3. It's because $$\mathcal B$$ is a basis of $$V$$, thus in particular every vector is a linear combination of them, and $$T$$ is linear.

4. $$T(v_i)=0$$ since $$v_i\in\ker T$$.

• It's by construction: first let $W$ be an arbitrary basis of $V$, then consider an arbitrary basis $U$ of $\ker T$. $U$ remains linearly independent in $V$. Apply Steinitz exchange lemma to obtain $W'\subseteq W$ with the given property. Commented Dec 27, 2020 at 19:58

let $$v_1,v_2,...,v_n$$ be a basis for $$V$$ then for all $$v \in V$$

$$v=c_1v_1+...+c_nv_n$$ $$\Rightarrow$$ $$T(v)=T(C_1v_1+...+c_nv_n)=c_1T(v_1)+...+c_nT(v_n) \in \text{span} \;T (\mathcal B)$$

similarly, $$w \in \text{span} \;T (\mathcal B) \Rightarrow w=T(c_1v_1+...+c_nv_n) \in Im(T)$$

that is why $$Im(T)=\text{span} \;T (\mathcal B)$$

• yes these are the hidden steps that were not written in my proof Commented Dec 27, 2020 at 17:52